A 1.40 F capacitor is charged through a 123 resistor and then discharged through

Question

A 1.40 F capacitor is charged through a 123 resistor and then discharged through the same resistor by short-circuiting the battery?

A) While the capacitor is being charged, find the time for the charge on its plates to reach 1e of its maximum value.

B) While the capacitor is being charged, find the current in the circuit at the time when the charge on its plates has reached 1e of its maximum value.

C) During the discharge of the capacitor, find the time for the charge on its plates to decrease to 1e of its initial value.

D) Find the time for the current in the circuit to decrease to 1e  of its initial value.

Explanation / Answer

Time constant of the ckt, T = R*C

= 123*1.4*10^-6

= 172.2*10^-6 s

A) charge on capacitor plates at time t,

Q = Qmax*(1 - e^(-t/T))

(1/e)*Qmax = Qmax*(1 - e^(-t/T))

1/e = 1 - e^(-t/T)

e^(-t/T) = 1 - 1/e

t = -T*ln(1-1/e)

= -172.2*10^-6*ln(1- 1/e)

= 7.9*10^-5 s

B) during charging,
I = Imax*e^(-t/T)

(1/e)*Imax = Imax*e^(-t/T)

1/e = e^(-t/T)

e^(-t/T) =1/e

t = -T*ln(1/e)

= -172.2*10^-6*ln(1/e)

= 1.72*10^-4 s

C) during discharge

I = Imax*e^(-t/T)

(1/e)*Qmax = Qmax*e^(-t/T)

1/e = e^(-t/T)

e^(-t/T) =1/e

t = -T*ln(1/e)

= -172.2*10^-6*ln(1/e)

= 1.72*10^-4 s

D)

during discharging,
I = Imax*e^(-t/T)

(1/e)*Imax = Imax*e^(-t/T)

1/e = e^(-t/T)

e^(-t/T) =1/e

t = -T*ln(1/e)

= -172.2*10^-6*ln(1/e)

= 1.72*10^-4 s

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